Q 12-09-191JEE MainJEE Main 2020 (5 Sep, Shift 1)Medium
A compound microscope consists of an objective lens of focal length $1\ \text{cm}$ and an eye piece of focal length $5\ \text{cm}$ with a separation of $10\ \text{cm}$. The distance between an object and the objective lens, at which the strain on the eye is minimum is $\dfrac{n}{40}$ cm. The value of $n$ is ______.
Numerical value type. Enter your answer.
Answer: 50
Minimum strain: final image at infinity, so the intermediate image is at the focus of the eyepiece, $5$ cm from it, i.e. $v = 10 - 5 = 5$ cm from the objective.
$$\frac1v - \frac1u = \frac1{f_o} \Rightarrow \frac15 - \frac1u = 1 \Rightarrow u = -\frac54\ \text{cm} = -\frac{50}{40}\ \text{cm}$$
So $n = 50$.
Solution by Sreeraj P, M.Sc Physics